I need a bit of help. I'm trying to formally prove why the sum of all of number's roots equals zero. I can logic it out in a way that makes sense to my brain, but I'm not sure how to translate that to proper math. I'm assuming that "A circle is close enough to 0" isn't a valid mathematical statement, of course.
So I start with the root equation, n√(r)cis((θ+k360°)/n). n√(r) will cancel out in the end because it's multiplying at the end for all possible roots, so that can be ignored. Then something something arccis(coolest operation ever) on both sides so the cis cancels out, and then you're left Σ(θ+k360°)/n = 0 and this is where I get unsure of what the fuck to do. My first instinct is to say that the /n can be ignored because the rule that says a fraction has to have a numerator of 0 to = 0, but the summation makes me unsure.
There are a few simplifications you can do without loss of generality. Firstly, remember that each root is spaced equally apart with respect to the angle in the real-imaginary plane. This means that if
n is even, there will always be a root with an opposite real-imaginary vector and thus cancel each other out. Thus you've proved half the problem.
Secondly, notice that the
θ is constant and constitutes nothing more than a translation (or rotation in the real-imaginary plane). Thus you only need to look at the case with
θ = 0 and odd
n.
This essentially means all roots are solutions to the equation
xn = 1, for odd
n. You will always have one root
x = 1. This is the case
k = 0. It also means all the complex roots are complex conjugates and thus the imaginary vector components cancel each other out.
So really, all you have to prove is Σcos(2
kπ/
n) = -1, for
k = 1 to
k =
n-1 and
n is odd, or equivalently Σcos(2
kπ/
n) = -1/2 for
k = 1 to
k = (
n-1)/2 and
n is odd, or equivalently Σcos(2
kπ/
n) = 0 for
k = 0 to
k =
n-1 and
n is odd.
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Topic: Mathematics Help Thread (Read 228657 times)